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AS 18

Carboxylic acids and derivatives

AS Level · Organic chemistry · Papers 1, 2 and 3 · extended in A2 33

🎯What you need to be able to do

  • Recall three ways of making carboxylic acids: oxidation of primary alcohols and aldehydes, hydrolysis of nitriles, and hydrolysis of esters.
  • Describe the reactions of carboxylic acids with reactive metals, alkalis and carbonates.
  • Describe esterification with alcohols, and reduction by LiAlH4 to a primary alcohol.
  • Recall how esters are made, and describe their hydrolysis by dilute acid and by dilute alkali.

📚The chemistry

The carboxyl group, –COOH, is a carbonyl and a hydroxyl on the same carbon. The two affect each other: the C=O pulls electron density away from the O–H, so the O–H is far more acidic than in an alcohol. Carboxylic acids are weak acids — only partly dissociated in water — but strong enough to react with carbonates, which alcohols cannot do.

\[ \mathrm{CH_3COOH(aq) \rightleftharpoons CH_3COO^{-}(aq) + H^{+}(aq)} \]

The anion is named with -oate: ethanoate, propanoate. Its salts are sodium ethanoate, calcium propanoate and so on.

18.1 Making carboxylic acids

  • Oxidation of a primary alcohol or an aldehyde — acidified K2Cr2O7 or KMnO4, heated under reflux (topic 16). \[ \mathrm{CH_3CH_2OH + 2[O] \rightarrow CH_3COOH + H_2O} \] \[ \mathrm{CH_3CHO + [O] \rightarrow CH_3COOH} \]
  • Hydrolysis of a nitrile — heat under reflux with dilute acid, or with dilute alkali followed by acidification (topic 19). \[ \mathrm{CH_3CH_2CN + 2H_2O + HCl \rightarrow CH_3CH_2COOH + NH_4Cl} \]
  • Hydrolysis of an ester — heat with dilute acid, or with dilute alkali followed by acidification (see 18.2).

“Followed by acidification” matters whenever alkali is used: in alkaline solution the acid exists as its salt (the carboxylate ion). Adding a strong acid such as dilute HCl at the end converts RCOO− back into RCOOH.

Reactions of carboxylic acids

They behave as typical (weak) acids:

  • With reactive metals — a redox reaction giving a salt and hydrogen; the metal slowly fizzes and dissolves. \[ \mathrm{2CH_3COOH + Mg \rightarrow (CH_3COO)_2Mg + H_2} \]
  • With alkalis — neutralisation, giving a salt and water. \[ \mathrm{CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O} \]
  • With carbonates — an acid–base reaction giving a salt, water and carbon dioxide (effervescence). This distinguishes a carboxylic acid from an alcohol or a phenol, none of which fizz with carbonates. \[ \mathrm{2CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + H_2O + CO_2} \]
  • Esterification — heat with an alcohol and a few drops of concentrated H2SO4 as catalyst (below).
  • Reduction by LiAlH4 (in dry ether) to a primary alcohol. NaBH4 is not strong enough. \[ \mathrm{CH_3COOH + 4[H] \rightarrow CH_3CH_2OH + H_2O} \]

18.2 Esters

Making esters

An ester forms in a condensation reaction between a carboxylic acid and an alcohol, with concentrated H2SO4 as catalyst, on heating. The reaction is reversible, so the yield is limited by the equilibrium:

\[ \mathrm{CH_3COOH + CH_3CH_2OH \rightleftharpoons CH_3COOCH_2CH_3 + H_2O} \]

Naming: the alcohol part first (as an alkyl group), the acid part second (as -oate). The ester above is ethyl ethanoate. In the formula, though, the acid part is usually written first: CH3COO–CH2CH3. The C=O always belongs to the acid part — use that to split a formula correctly.

Esters have sweet, fruity smells and are used as flavourings, perfumes and solvents.

Hydrolysing esters

Hydrolysis splits the ester back into the acid and the alcohol. There are two ways, and the difference between them is examined:

  • Dilute acid, heat under reflux. The reverse of esterification, so it is reversible and does not go to completion. The acid is a catalyst. \[ \mathrm{CH_3COOCH_2CH_3 + H_2O \rightleftharpoons CH_3COOH + CH_3CH_2OH} \]
  • Dilute alkali (NaOH), heat under reflux. The acid formed is immediately converted into its salt, which cannot react with the alcohol, so the reaction goes to completion. The products are the carboxylate salt and the alcohol; adding dilute acid afterwards gives the free carboxylic acid. \[ \mathrm{CH_3COOCH_2CH_3 + NaOH \rightarrow CH_3COONa + CH_3CH_2OH} \]

Alkaline hydrolysis is therefore the better way to hydrolyse an ester completely. (Alkaline hydrolysis of fats, which are esters, is how soap is made.)

✏️Worked example

0.370 g of a carboxylic acid, RCOOH, is dissolved in water and titrated with 0.200 mol dm−3 sodium hydroxide; 25.0 cm3 is needed. (a) Calculate the Mr of the acid and identify it. (b) The acid is heated with methanol and concentrated sulfuric acid. Name the ester and write the equation. (c) Compare the products of hydrolysing this ester with (i) dilute hydrochloric acid and (ii) aqueous sodium hydroxide. [Ar: C 12.0, H 1.0, O 16.0]

(a) RCOOH + NaOH → RCOONa + H2O, a 1 : 1 ratio.

\[ n(\mathrm{NaOH}) = 0.200 \times \frac{25.0}{1000} = 5.00 \times 10^{-3}\ \mathrm{mol} = n(\mathrm{RCOOH}) \] \[ M_\mathrm{r} = \frac{0.370}{5.00 \times 10^{-3}} = 74.0 \]

COOH has mass 45.0, so R = 74.0 − 45.0 = 29.0, which is C2H5. The acid is propanoic acid, CH3CH2COOH.

(b) Methyl propanoate:

\[ \mathrm{CH_3CH_2COOH + CH_3OH \rightleftharpoons CH_3CH_2COOCH_3 + H_2O} \]

(c)(i) With dilute HCl, heated under reflux: an equilibrium mixture containing propanoic acid and methanol, with unreacted ester, because acid hydrolysis is reversible. (ii) With NaOH(aq), heated under reflux: sodium propanoate and methanol, and the reaction goes to completion, because the propanoate ion cannot re-form the ester. Propanoic acid itself is only obtained after acidifying.

Check it. The formula C2H5COOH gives 3(12.0) + 6(1.0) + 2(16.0) = 74.0, matching the titration exactly. The only other C3H6O2 compounds are esters (methyl ethanoate, ethyl methanoate), which would not react 1 : 1 with NaOH in a quick titration at room temperature.
Naming the ester backwards. CH3CH2COOCH3 is methyl propanoate, not “propyl methanoate”. Find the C=O: the fragment containing it (CH3CH2CO–) is the acid, and the alkyl group on the other oxygen is the alcohol. A second trap in (c): giving “propanoic acid” as the product of alkaline hydrolysis. In alkali you get the salt.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Give three different starting materials that could be converted into butanoic acid, with the reagent for each.
Butan-1-ol (or butanal): acidified potassium dichromate(VI), heat under reflux. Butanenitrile, CH3CH2CH2CN: dilute acid, heat under reflux (or dilute NaOH then acidify). An ester of butanoic acid, such as ethyl butanoate: dilute acid and heat, or dilute NaOH and heat followed by acidification.
2. Describe a simple test that distinguishes ethanoic acid from ethanol, with observations and an equation.
Add sodium carbonate (or sodium hydrogencarbonate) solution. Ethanoic acid gives effervescence; the gas turns limewater milky. 2CH3COOH + Na2CO3 → 2CH3COONa + H2O + CO2. Ethanol gives no reaction, because it is far too weak an acid. (Sodium metal would not distinguish them: both fizz.)
3. Write equations for the reaction of propanoic acid with (a) zinc; (b) potassium hydroxide. Classify each reaction.
(a) 2CH3CH2COOH + Zn → (CH3CH2COO)2Zn + H2: a redox reaction (Zn 0 → +2, H +1 → 0), giving zinc propanoate and hydrogen. (b) CH3CH2COOH + KOH → CH3CH2COOK + H2O: neutralisation (acid–base), giving potassium propanoate and water.
4. Name the ester CH3COOCH2CH2CH3, and give the acid and alcohol it is made from.
The C=O is in the CH3CO– fragment, so the acid part is ethanoate (from ethanoic acid); the group on the other oxygen, CH2CH2CH3, is propyl (from propan-1-ol). The ester is propyl ethanoate (it smells of pears).
5. Explain why hydrolysis of an ester with sodium hydroxide goes to completion, but hydrolysis with dilute acid does not.
Acid hydrolysis is simply the reverse of esterification: RCOOR′ + H2O ⇌ RCOOH + R′OH. The carboxylic acid and alcohol produced can react to re-form the ester, so an equilibrium is set up. In alkaline hydrolysis the carboxylic acid is converted into its carboxylate salt as soon as it forms (RCOOH + OH− → RCOO− + H2O). The carboxylate ion does not react with the alcohol, so the reverse reaction cannot occur, and all the ester is hydrolysed.
6. Ethanoic acid is reduced with LiAlH4. Name the product, write an equation, and explain why NaBH4 cannot be used.
Product: ethanol. CH3COOH + 4[H] → CH3CH2OH + H2O. The carboxyl group is much harder to reduce than the carbonyl group of an aldehyde or ketone; only the more powerful reducing agent LiAlH4 (in dry ether) can do it. NaBH4 is a milder reducing agent that reduces aldehydes and ketones only.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Chemguide (Jim Clark) — the carboxylic acids and esters pages, including the hydrolysis of esters in acid and alkali
  • Royal Society of Chemistry — practical guide for making an ester and identifying it by smell